Which triangles are similar?

Which function represents a horizontal

mote1985 [20]

Answer:

y=2^3x

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

xion orders 5 loves of bread from the website bakery the total shipping weight is 9 pounds as model of the bread what is in poun

Ray Of Light [21]

Each loaf of bread is 1.8 pounds

6 0

2 years ago

Find integral∫cos³x dx) )?

son4ous [18]

Answer:

sin(x)−1/3(sin3(x))+C

Step-by-step explanation:

use u substitution method to integrate

4 0

2 years ago

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Un arquitecto ha pagado en un corralón un total de $ 130 por 20 Ladrillos, 5 kg de Arena y 10 kg de Cemento. Calcular el precio

Advocard [28]

Answer:

1 ladrillo cuesta 1 dólar. 1 brick costs 1 dollars.

1 kg de cemento cuesta 3 dólares: 1 kg cement costs 3 dollars

1 kg de arena cuesta $ 16: 1 kg sand costs $ 16

Step-by-step explanation:

Spanish

Deje que los ladrillos, la arena y el cemento se denoten por b, sy c. Luego, de acuerdo con la condición dada

20b + 5s + 10c = $ 130 --- ecuación 1

1c = 3b ------ ecuación 2

1s = 4c + 4b ------ ecuación 3

Poniendo la ecuación 2 en la ecuación 3

1s = 4 (3b) + 4b

1s = 16b ---------- ecuación 4

20b + 5s + 10 (3b) = $ 130

50b + 5s = 130 --- ecuación 5

Poniendo la ecuación 4 en la ecuación 5

50 b + 5 (16b) = 130

50 b + 80b = 130

130 b = 130

b = 130/130 = $ 1 ------- ecuación a

1 ladrillo cuesta 1 dólar.

Poner el valor de costo de 1 ladrillo en estas ecuaciones

1c = 3b ------ ecuación 2

1s = 4c + 4b ------ ecuación 3

1c = 3 (1) = $ 3 -------- ecuación b

1 kg de cemento cuesta 3 dólares

1s = 4c + 4b

1 s = 4 (3) + 4 (1)

1 s = 12 + 4

1s = $ 16 ------ ecuación c

1 kg de arena cuesta $ 16

Poner valores de la ecuación a, b, c en la ecuación 1

20b + 5s + 10c = $ 130

20 (1) + 5 (16) + 10 (3) = 130

20 + 80 + 30 = 130

130 = 130 ------------ por lo tanto probado

English

Let the bricks , sand and cement be denoted by b, s and c. Then according to the given condition

20b+ 5s+ 10c= $130--- equation 1

1c= 3b------ equation 2

1s= 4c+4b------ equation 3

Putting equation 2 in equation 3

1s= 4(3b) + 4b

1s= 16b---------- equation 4

20b+ 5s+ 10(3b)= $130

50b + 5s= 130--- equation 5

Putting equation 4 in equation 5

50 b + 5 (16b) = 130

50 b+ 80b= 130

130 b= 130

b= 130/130= $1------- equation a

1 brick costs 1 dollars.

Putting the cost value of 1 brick in these equations

1c= 3b------ equation 2

1s= 4c+4b------ equation 3

1c= 3(1)= $3-------- equation b

1 kg cement costs 3 dollars

1s= 4c+4b

1s= 4(3) + 4(1)

1s= 12+4

1s= $16 ------ equation c

1 kg sand costs $ 16

Putting values from equation a, b, c in equation 1

20b+ 5s+ 10c= $130

20 (1) + 5 (16) + 10 (3)= 130

20 + 80 + 30 = 130

130 = 130 ------------ hence proved

8 0

2 years ago

A recent health report revealed that a woman with insurance spends an average of 2.3 days in the hospital following a routine ch

cupoosta [38]

Answer:

$t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385$

$p_v =2*P(t_{30}>2.385)=0.0236$

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

$(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611$

$(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861$

Step-by-step explanation:

Data given and notation

$\bar X_{insurance}=2.3$ represent the mean for insurance

$\bar X_{No. ins}=1.9$ represent the mean withour insurance

$s_{Insurance}=0.6$ represent the sample standard deviation for the insurance case

$s_{No. Ins}=0.3$ represent the sample standard deviation for the No insurance case

$n_{Insurance}=16$ sample size for insurance

$n_{No. Iss}=16$ sample size for no insurance

t would represent the statistic (variable of interest)

$\alpha=0.01$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis: $\mu_{Insu}=\mu_{No. Ins}$

Alternative hypothesis: $\mu_{Ins} \neq \mu_{No. Ins}$

Since we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

$t=\frac{\bar X_{Ins}-\bar X_{No.Ins}}{\sqrt{\frac{s^2_{Ins}}{n_{Ins}}+\frac{s^2_{No. Ins}}{n_{No. Ins}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385$

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

$df = n_{ins} +n_{No Ins}-2=16+16-2 =30$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{30}>2.385)=0.0236$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

$(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611$

$(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861$

6 0

2 years ago